Question

How much oxygen is formed from 264g of co2 and 216g of h2o?

Answer 1

Answer : The amount of oxygen formed, 576 grams

Solution : Given,

Mass of $CO_2$ = 264 g

Mass of $H_2O$ = 216 g

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $CO_2$ and $H_2O$.

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{264g}{44g/mole}=6moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{216g}{18g/mole}=12moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2$

From the balanced reaction we conclude that

As, 6 moles of $CO_2$ react with 6 moles of $H_2O$

So, 6 moles of $CO_2$ react with $\frac{6}{6}\times 6=6$ moles of $H_2O$

That means, in the given balanced reaction, $CO_2$ is a limiting reagent because it limits the formation of products and $H_2O$ is an excess reagent.

The excess reagent remains $(H_2O)$  = 12 - 6 = 6 moles

Now we have to calculate the moles of $O_2$.

As, 2 moles of $CO_2$ react with 6 moles of $O_2$

So, 6 moles of $CO_2$ react with $\frac{6}{2}\times 6=18$ moles of $O_2$

Now we have to calculate the mass of $O_2$.

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(18mole)\times (32g/mole)=576g$

Therefore, the amount of oxygen formed, 576 grams